Question: Simplify; express your answer in exponential form. Assume $y\neq 0, r\neq 0$. $\dfrac{{(y^{-2})^{-2}}}{{(y^{5}r^{4})^{-5}}}$
Answer: To start, try working on the numerator and the denominator independently. In the numerator, we have ${y^{-2}}$ to the exponent ${-2}$ . Now ${-2 \times -2 = 4}$ , so ${(y^{-2})^{-2} = y^{4}}$ In the denominator, we can use the distributive property of exponents. ${(y^{5}r^{4})^{-5} = (y^{5})^{-5}(r^{4})^{-5}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(y^{-2})^{-2}}}{{(y^{5}r^{4})^{-5}}} = \dfrac{{y^{4}}}{{y^{-25}r^{-20}}}$ Break up the equation by variable and simplify. $\dfrac{{y^{4}}}{{y^{-25}r^{-20}}} = \dfrac{{y^{4}}}{{y^{-25}}} \cdot \dfrac{{1}}{{r^{-20}}} = y^{{4} - {(-25)}} \cdot r^{- {(-20)}} = y^{29}r^{20}$.